Inequality Proof Problems [201-250]
Recommended posting: 【Algebra】 Algebra Index
Restructured the IneqMath training data.
P201. Let $a, b, c \in \mathbb{R}^{+}$ such that $ab + bc + ca = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[\frac{1}{a(a+b)}+\frac{1}{b(b+c)}+\frac{1}{c(c+a)} \geq C\]S201. $C = \frac{9}{2}$
The given inequality is equivalent to
\[\frac{c(a+b)+ab}{a(a+b)}+\frac{a(b+c)+bc}{b(b+c)}+\frac{b(c+a)+ac}{c(c+a)} \ge \frac{9}{2},\]i.e.
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\ge \frac{9}{2}.\]Equivalently,
\[\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a} +\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a} \ge \frac{15}{2}. \qquad (1)\]We have
\[\begin{aligned} &\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a} +\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\\ &=\left(\frac{a+b}{4b}+\frac{b+c}{4c}+\frac{c+a}{4a} +\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\right) +\frac{3}{4}\left(\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}\right)\\ &\ge 6\sqrt[6]{\frac{a+b}{4b}\cdot\frac{b+c}{4c}\cdot\frac{c+a}{4a} \cdot\frac{b}{a+b}\cdot\frac{c}{b+c}\cdot\frac{a}{c+a}} +\frac{3}{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\right)\\ &=6\sqrt[6]{\frac{1}{64}}+\frac{3}{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\right)\\ &=3+\frac{3}{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\right)\\ &\ge 3+\frac{3}{4}\left(3\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}+3\right) =3+\frac{3}{4}(3+3)=\frac{15}{2}, \end{aligned}\]as required.
Equality holds when $(a=b=c=\frac{1}{\sqrt{3}})$, since $(ab+bc+ca=3\left(\frac{1}{\sqrt{3}}\right)^{2}=1)$, and all steps above become equalities. Therefore, the largest constant is $(C=\frac{9}{2})$.
P202. Let $a, b, c$ be positive real numbers. Find the smallest positive number $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[\frac{a+\sqrt{a b}+\sqrt[3]{a b c}}{C} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}\]S202. $C = 3$
Applying $(AM \ge GM)$ we get
\[\sqrt[3]{ab\cdot \frac{a+b}{2}} \ge \sqrt[3]{ab\cdot \sqrt{ab}} = \sqrt{ab}.\]So
\[a+\sqrt{ab}+\sqrt[3]{abc} \le a+\sqrt[3]{ab\cdot \frac{a+b}{2}}+\sqrt[3]{abc}.\]Now it is enough to show that
\[a+\sqrt[3]{ab\cdot \frac{a+b}{2}}+\sqrt[3]{abc} \le 3\sqrt[3]{a\cdot \frac{a+b}{2}\cdot \frac{a+b+c}{3}}.\]Another application of $(AM \ge GM)$ gives
\[\sqrt[3]{1\cdot \frac{2a}{a+b}\cdot \frac{3a}{a+b+c}} \le \frac{1+\frac{2a}{a+b}+\frac{3a}{a+b+c}}{3},\] \[\sqrt[3]{1\cdot 1\cdot \frac{3b}{a+b+c}} \le \frac{2+\frac{3b}{a+b+c}}{3},\]and
\[\sqrt[3]{1\cdot \frac{2b}{a+b}\cdot \frac{3c}{a+b+c}} \le \frac{1+\frac{2b}{a+b}+\frac{3c}{a+b+c}}{3}.\]Adding, we obtain
\[\sqrt[3]{\frac{2a}{a+b}\cdot \frac{3a}{a+b+c}} +\sqrt[3]{\frac{3b}{a+b+c}} +\sqrt[3]{\frac{2b}{a+b}\cdot \frac{3c}{a+b+c}} \le 3.\]i.e.
\[\sqrt[3]{\frac{1}{a}\cdot \frac{2}{a+b}\cdot \frac{3}{a+b+c}} \left(a+\sqrt[3]{ab\cdot \frac{a+b}{2}}+\sqrt[3]{abc}\right) \le 3,\]i.e.
\[a+\sqrt[3]{ab\cdot \frac{a+b}{2}}+\sqrt[3]{abc} \le 3\sqrt[3]{a\cdot \frac{a+b}{2}\cdot \frac{a+b+c}{3}}.\]Equality holds when $(a=b=c)$. Therefore, the answer is $(C=3)$.
P203. Let $a, b, c, d, e$ be positive real numbers such that $a+b+c+d+e=5$. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c, d, e$ satisfying the given constraint:
\[a b c + b c d + c d e + d e a + e a b \leq C.\]S203. $C = 5$
Without loss of generality, we may assume that
\[e=\min\{a,b,c,d,e\}.\]By $(AM\ge GM)$, we have
\[\begin{aligned} abc+bcd+cde+dea+eab &= e(a+c)(b+d)+bc(a+d-e)\\ &\le e\left(\frac{a+c+b+d}{2}\right)^{2} +\left(\frac{b+c+a+d-e}{3}\right)^{3}\\ &= \frac{e(5-e)^{2}}{4}+\frac{(5-2e)^{3}}{27}. \end{aligned}\]So it suffices to prove that
\[\frac{e(5-e)^{2}}{4}+\frac{(5-2e)^{3}}{27}\le 5,\]which can be rewritten as
\[(e-1)^{2}(e+8)\ge 0,\]which is obviously true.
Equality holds if and only if $(a=b=c=d=e=1)$. In this case,
\[abc+bcd+cde+dea+eab = 5.\]Therefore, the minimal constant is $(C=5)$.
P204. Let $a, b, c$ be non-negative real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}_0$:
\[a^{3}+b^{3}+c^{3}+a b c \geq C(a+b+c)^{3}\]S204. $C = \frac{1}{7}$
We have
\[\begin{aligned} (a+b+c)^{3} &=a^{3}+b^{3}+c^{3}+3\big(a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\big)+6abc\\ &=\frac{T[3,0,0]}{2}+3T[2,1,0]+T[1,1,1]. \end{aligned}\]and
\[a^{3}+b^{3}+c^{3}+abc=\frac{T[3,0,0]}{2}+\frac{T[1,1,1]}{6}.\]So we need to prove that
\[7\left(\frac{T[3,0,0]}{2}+\frac{T[1,1,1]}{6}\right) \ge \frac{T[3,0,0]}{2}+3T[2,1,0]+T[1,1,1].\]i.e.
\[3T[3,0,0]+\frac{T[1,1,1]}{6}\ge 3T[2,1,0],\]which is true because $(T[3,0,0]\ge T[2,1,0])$ and $(T[1,1,1]\ge 0)$ (Muirhead’s theorem).
Equality holds when $(a=b=c)$. Therefore, the answer is
\[C=\frac{1}{7}.\]P205. Let $a, b, c, d > 1$ be real numbers. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c, d > 1$:
\[\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1} \leq C \sqrt{(a b+1)(c d+1)}.\]S205. $C = 1$
We will prove that for every $(x,y\in\mathbb{R}^+)$ we have
\[\sqrt{x-1}+\sqrt{y-1}\le \sqrt{xy}.\]Applying the Cauchy–Schwarz inequality to $((\sqrt{x-1},\,1)) and ((1,\,\sqrt{y-1}))$ gives
\[(\sqrt{x-1}+\sqrt{y-1})^{2} \le \big((x-1)+1\big)\big(1+(y-1)\big) =xy,\]hence
\[\sqrt{x-1}+\sqrt{y-1}\le \sqrt{xy}.\]Now we deduce that
\[\sqrt{a-1}+\sqrt{b-1}\le \sqrt{ab}, \qquad \sqrt{c-1}+\sqrt{d-1}\le \sqrt{cd},\]and therefore
\[\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1} \le \sqrt{ab}+\sqrt{cd}.\]Also, applying the same inequality with $(x=ab+1)$ and $(y=cd+1)$ yields
\[\sqrt{ab}+\sqrt{cd} =\sqrt{(ab+1)-1}+\sqrt{(cd+1)-1} \le \sqrt{(ab+1)(cd+1)}.\]Combining these, we obtain
\[\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1} \le \sqrt{(ab+1)(cd+1)}.\]Equality holds when $(a=b=c=d=1)$, in which case both sides are $(0)$. Therefore, the sharp constant is $(C=1)$.
P206. Let $n > 2$ and let $x_{1}, x_{2}, \ldots, x_{n} > 0$ such that $\sum_{i=1}^{n} x_i = 1$. Find the maximal constant $C$ such that the following inequality holds for all $x_{i}$:
\[\prod_{i=1}^{n}\left(1+\frac{1}{x_{i}}\right) \geq C \prod_{i=1}^{n}\left(\frac{n-x_{i}}{1-x_{i}}\right).\]S206. $C =1$
The most natural idea is to use the fact that
\[\frac{n-x_i}{1-x_i} =1+\frac{n-1}{x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n}.\]Thus, we have
\[\prod_{i=1}^{n}\left(\frac{n-x_i}{1-x_i}\right) \le \prod_{i=1}^{n}\left(1+\frac{1}{\sqrt[n-1]{x_1x_2\cdots x_{i-1}x_{i+1}\cdots x_n}}\right),\]and we have to prove the inequality
\[\prod_{i=1}^{n}\left(1+\frac{1}{x_i}\right) \ge \prod_{i=1}^{n}\left(1+\frac{1}{\sqrt[n-1]{x_1x_2\cdots x_{i-1}x_{i+1}\cdots x_n}}\right).\]But this one is not very hard, because it follows immediately by multiplying the inequalities
\[\prod_{j\ne i}\left(1+\frac{1}{x_j}\right) \ge \left(1+\sqrt[n-1]{\prod_{j\ne i}\frac{1}{x_j}}\right)^{\,n-1},\]obtained from Huygens’ inequality.
We will prove even more, namely that
\[\prod_{i=1}^{n}\left(1+\frac{1}{x_i}\right) \ge \left(\frac{n^{2}-1}{n}\right)^{n}\cdot \prod_{i=1}^{n}\frac{1}{1-x_i}.\]It is clear that this inequality is stronger than the initial one.
First, let us prove that
\[\prod_{i=1}^{n}\frac{1+x_i}{1-x_i} \ge \left(\frac{n+1}{n-1}\right)^{n}.\]This follows from Jensen’s inequality for the convex function $(f(x)=\ln(1+x)-\ln(1-x))$.
So, it suffices to prove that
\[\frac{\left(\frac{n+1}{n-1}\right)^{n}}{\prod_{i=1}^{n}x_i}\cdot \prod_{i=1}^{n}(1-x_i)^{2} \ge \left(\frac{n^{2}-1}{n}\right)^{n}.\]But a quick look shows that this is exactly the inequality proved in the solution of Problem 121.
Equality holds when $(x_1=x_2=\cdots=x_n=\frac{1}{n})$, since then all terms are equal and the inequalities used above become equalities. Therefore, the answer is $(C=1)$.
P207. Let $a, b, c \geq 0$ such that $a + b + c = 1$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$:
\[a^{2} + b^{2} + c^{2} \leq C(a^{3} + b^{3} + c^{3}) + 3abc\]S207. $C = 2$
Let $(p=a+b+c=1)$, $(q=ab+bc+ca)$, $(r=abc)$.
The given inequality is equivalent to
\[1-2q \le 2(1-3q+3r)+3r \quad\Longleftrightarrow\quad 4q \le 1+9r.\]For $((a,b,c)=(1,0,0))$ and its permutations we have $(q=0)$ and $(r=0)$, so the inequality becomes
\[0 \le 1,\]which is satisfied.
Therefore, the answer is $(C=2)$.
P208. Let $a, b, c, d > 0$ be real numbers such that $a \leq 1$, $a+b \leq 5$, $a+b+c \leq 14$, and $a+b+c+d \leq 30$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraints:
\[\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \leq C\]S208. $C = 10$
The function $(f:(0,+\infty)\to(0,+\infty))$ defined by $(f(x)=\sqrt{x})$ is concave on $((0,+\infty))$, so by Jensen’s inequality, for
\[n=4,\qquad \alpha_1=\frac{1}{10},\ \alpha_2=\frac{2}{10},\ \alpha_3=\frac{3}{10},\ \alpha_4=\frac{4}{10},\]we get
\[\frac{1}{10}\sqrt{a} +\frac{2}{10}\sqrt{\frac{b}{4}} +\frac{3}{10}\sqrt{\frac{c}{9}} +\frac{4}{10}\sqrt{\frac{d}{16}} \le \sqrt{\frac{a}{10}+\frac{b}{20}+\frac{c}{30}+\frac{d}{40}}.\]i.e.
\[\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \le 10\sqrt{\frac{12a+6b+4c+3d}{120}}. \qquad (1)\]On the other hand, we have
\[\begin{aligned} 12a+6b+4c+3d &=3(a+b+c+d)+(a+b+c)+2(a+b)+6a\\ &\le 3\cdot 30+14+2\cdot 5+6\cdot 1=120. \end{aligned}\]By (1) and the last inequality we obtain the required result.
Equality holds when $(a=1), (b=4), (c=9), (d=16)$, since these values satisfy all constraints with equality and yield
\[\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}=1+2+3+4=10.\]Therefore, the answer is $(C=10)$.
P209. Prove that if $n \geq 2$ and $a_{1}, a_{2}, \ldots, a_{n}$ are real numbers with product 1, then find the largest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$:
\[a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}-n \geq C \cdot\frac{n}{n-1} \cdot \sqrt[n]{n-1}\left(a_{1}+a_{2}+\cdots+a_{n}-n\right)\]S209. $C = 2$
We will prove the inequality by induction. For $(n=2)$ it is trivial. Now suppose the inequality is true for $(n-1)$ numbers, and let us prove it for $(n)$.
First, it is enough to prove it for $(a_1,\ldots,a_n>0)$ (otherwise we replace $(a_1,a_2,\ldots,a_n)$ with $(\left|a_1\right|,\left|a_2\right|,\ldots,\left|a_n\right|)$, which have product $(1)$, while the right-hand side increases).
Let $(a_n)$ be the maximum among $(a_1,a_2,\ldots,a_n)$, and let $(G)$ be the geometric mean of $(a_1,a_2,\ldots,a_{n-1})$. First, we will prove that
\[\begin{aligned} &a_1^{2}+a_2^{2}+\cdots+a_n^{2}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\,(a_1+a_2+\cdots+a_n-n)\\ &\ge a_n^{2}+(n-1)G^{2}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\,(a_n+(n-1)G-n). \end{aligned}\]This is equivalent to
\[a_1^{2}+\cdots+a_{n-1}^{2}-(n-1)\sqrt[n-1]{a_1^{2}a_2^{2}\cdots a_{n-1}^{2}} \ge \frac{2n}{n-1}\sqrt[n]{n-1}\left(a_1+\cdots+a_{n-1}-(n-1)\sqrt[n-1]{a_1a_2\cdots a_{n-1}}\right).\]Because $(\sqrt[n-1]{a_1a_2\cdots a_{n-1}}\le 1)$ and
\[a_1+\cdots+a_{n-1}-(n-1)\sqrt[n-1]{a_1a_2\cdots a_{n-1}}\ge 0,\]it is enough to prove
\[a_1^{2}+\cdots+a_{n-1}^{2}-(n-1)G^{2} \ge \frac{2n}{n-1}\sqrt[n]{n-1}\,G\,(a_1+\cdots+a_{n-1}-(n-1)G).\]Now we apply the inductive hypothesis to the numbers $(\frac{a_1}{G},\ldots,\frac{a_{n-1}}{G})$, which have product $(1)$, and infer that
\[\frac{a_1^{2}+\cdots+a_{n-1}^{2}}{G^{2}}-(n-1) \ge \frac{2(n-1)}{n-2}\sqrt[n-1]{n-2}\left(\frac{a_1+\cdots+a_{n-1}}{G}-(n-1)\right).\]So it suffices to prove that
\[\frac{2(n-1)}{n-2}\sqrt[n-1]{n-2}\,(a_1+\cdots+a_{n-1}-(n-1)G) \ge \frac{2n}{n-1}\sqrt[n]{n-1}\,(a_1+\cdots+a_{n-1}-(n-1)G),\]which is the same as
\[1+\frac{1}{n(n-2)} \ge \frac{\sqrt[n]{n-1}}{\sqrt[n-1]{n-2}}.\]This becomes
\[\left(1+\frac{1}{n(n-2)}\right)^{n(n-1)} \ge \frac{(n-1)^{n-1}}{(n-2)^{n}}.\]For $(n>4)$ it follows from
\[\left(1+\frac{1}{n(n-2)}\right)^{n(n-1)} > 2,\]and
\[\frac{(n-1)^{n-1}}{(n-2)^{n}} =\frac{1}{n-2}\left(1+\frac{1}{n-2}\right)\left(1+\frac{1}{n-2}\right)^{n-2} <\frac{e}{n-2}\left(1+\frac{1}{n-2}\right)<2.\]For $(n=3)$ and $(n=4)$ it is easy to check.
Thus, we have proved that
\[\begin{aligned} &a_1^{2}+a_2^{2}+\cdots+a_n^{2}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\,(a_1+a_2+\cdots+a_n-n)\\ &\ge a_n^{2}+(n-1)G^{2}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\,(a_n+(n-1)G-n). \end{aligned}\]Now it is enough to prove that
\[x^{2(n-1)}+\frac{n-1}{x^{2}}-n \ge \frac{2n}{n-1}\sqrt[n]{n-1}\left(x^{n-1}+\frac{n-1}{x}-n\right)\]for all $(x\ge 1)$ (we took $(x=\frac{1}{G})$).
Let us consider the function
\[f(x)=x^{2(n-1)}+\frac{n-1}{x^{2}}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\left(x^{n-1}+\frac{n-1}{x}-n\right).\]We have
\[f'(x)=2\cdot \frac{x^{n}-1}{x^{2}}\left[\frac{(n-1)(x^{n}+1)}{x}-n\sqrt[n]{n-1}\right]\ge 0,\]because
\[x^{n-1}+\frac{1}{x} =x^{n-1}+\frac{1}{(n-1)x}+\cdots+\frac{1}{(n-1)x} \ge n\sqrt[n]{\frac{1}{(n-1)^{\,n-1}}}.\]Thus $(f)$ is increasing, so $(f(x)\ge f(1)=0)$. This proves the inequality.
Equality holds when $(a_1=a_2=\cdots=a_n=1)$. Therefore, the answer is $(C=2)$.
P210. Let $a, b, c$ be positive real numbers. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}} \geq C \left( \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \right)\]S210-1. $C = 1$
By the Cauchy–Schwarz inequality we have
\[(a+b+c)\left(\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}}\right) \ge \left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)^{2}.\]So it is enough to prove that
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge a+b+c.\]But this follows immediately from the Cauchy–Schwarz inequality:
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \ge \frac{(a+b+c)^{2}}{a+b+c} = a+b+c.\]Equality holds when $(a=b=c)$. Therefore, the maximal constant is $(C=1)$.
S210-2. $C = 1$
We have
\[\frac{a^{3}}{b^{2}} \ge \frac{a^{2}}{b}+a-b \iff a^{3}+b^{3}\ge ab(a+b) \iff (a-b)^{2}(a+b)\ge 0,\]which is clearly true.
Writing the analogous inequalities and adding them up gives
\[\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}} \ge \left(\frac{a^{2}}{b}+a-b\right)+\left(\frac{b^{2}}{c}+b-c\right)+\left(\frac{c^{2}}{a}+c-a\right) = \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}.\]Equality holds when $(a=b=c)$. Therefore, the maximal constant is $(C=1)$.
P211. Let $a, b, c$ be positive real numbers such that $a b c \geq 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[a+b+c \geq C\left(\frac{1+a}{1+b}+\frac{1+b}{1+c}+\frac{1+c}{1+a}\right).\]S211. $C = 1$
We have
\[\begin{aligned} &a+b+c-\frac{1+a}{1+b}-\frac{1+b}{1+c}-\frac{1+c}{1+a}\\ &=(1+a)\left(1-\frac{1}{1+b}\right) +(1+b)\left(1-\frac{1}{1+c}\right) +(1+c)\left(1-\frac{1}{1+a}\right)-3\\ &=\frac{(1+a)b}{1+b}+\frac{(1+b)c}{1+c}+\frac{(1+c)a}{1+a}-3\\ &\ge 3\sqrt[3]{\frac{(1+a)b}{1+b}\cdot\frac{(1+b)c}{1+c}\cdot\frac{(1+c)a}{1+a}}-3\\ &=3\sqrt[3]{abc}-3\\ &\ge 0 \qquad (abc\ge 1). \end{aligned}\]Equality holds when $(a=b=c=1)$. Therefore, the maximal constant is $(C=1)$.
P212. Let $h_{a}, h_{b}$, and $h_{c}$ be the lengths of the altitudes, and $R$ and $r$ be the circumradius and inradius, respectively, of a given triangle. Determine the minimal constant $C$ such that the following inequality holds for all triangles:
\[h_{a}+h_{b}+h_{c} \leq 2R+Cr.\]S212. $C = 5$
Lemma 21.4 In an arbitrary triangle we have
\[ab+bc+ca = r^{2}+s^{2}+4rR \quad \text{and} \quad a^{2}+b^{2}+c^{2}=2\bigl(s^{2}-4Rr-r^{2}\bigr).\]Proof. We have
\[\begin{aligned} r^{2}+s^{2}+4rR &= \frac{P^{2}}{s^{2}} + s^{2} + \frac{abc}{P}\cdot \frac{P}{s} \\ &= \frac{(s-a)(s-b)(s-c)}{s} + s^{2} + \frac{abc}{s} \\ &= \frac{s^{3}-as^{2}-bs^{2}-cs^{2}+abs+bcs+cas-abc+s^{3}+abc}{s} \\ &= 2s^{2}-s(a+b+c)+ab+bc+ca \\ &= 2s^{2}-2s^{2}+ab+bc+ca \\ &= ab+bc+ca. \end{aligned}\]Hence
\[ab+bc+ca = r^{2}+s^{2}+4rR \qquad (1)\]Now by (1) we have
\[\begin{aligned} ab+bc+ca &= r^{2}+s^{2}+4rR \\ &= \frac12\left(2r^{2}+8rR+\frac{(a+b+c)^{2}}{2}\right) \\ &= \frac12\left(2r^{2}+8rR+\frac{a^{2}+b^{2}+c^{2}}{2}\right) + \frac{ab+bc+ca}{2}, \end{aligned}\]from which it follows that
\[ab+bc+ca = 2r^{2}+8rR+\frac{a^{2}+b^{2}+c^{2}}{2} \qquad (2)\]Now (1) and (2) yield
\[a^{2}+b^{2}+c^{2}=2\bigl(s^{2}-4Rr-r^{2}\bigr) \qquad (3)\]Without proof we will give the following lemma. In an arbitrary triangle we have
\[s^{2} \le 4R^{2}+4Rr+3r^{2} \qquad (4)\]In an arbitrary triangle we have $(a^{2}+b^{2}+c^{2} \le 8R^{2}+4r^{2})$. Proof. From (3) and (4) we have
\[\begin{aligned} a^{2}+b^{2}+c^{2} &= 2\bigl(s^{2}-4Rr-r^{2}\bigr) \\ &\le 2\bigl(4R^{2}+4Rr+3r^{2}-4Rr-r^{2}\bigr) \\ &= 8R^{2}+4r^{2}. \end{aligned}\]Hence
\[a^{2}+b^{2}+c^{2} \le 8R^{2}+4r^{2} \qquad (5)\]Now let us consider our problem. We have
\[\begin{aligned} 2R(h_a+h_b+h_c) &= 2R\left(\frac{2P}{a}+\frac{2P}{b}+\frac{2P}{c}\right) = 4PR\left(\frac1a+\frac1b+\frac1c\right) \\ &= 4PR\cdot \frac{ab+bc+ca}{abc} = ab+bc+ca \\ &\stackrel{(2)}{=} 2r^{2}+8rR+\frac{a^{2}+b^{2}+c^{2}}{2} \\ &\stackrel{(5)}{\le} 2r^{2}+8rR+\frac{8R^{2}+4r^{2}}{2} \\ &= 4R^{2}+8Rr+4r^{2}. \end{aligned}\]Therefore
\[R(h_a+h_b+h_c) \le 2R^{2}+4Rr+2r^{2}.\]Since $(r\le \frac{R}{2})$, we have $(2r^{2}\le Rr)$, hence
\[R(h_a+h_b+h_c)\le 2R^{2}+5Rr = R(2R+5r),\]so
\[h_a+h_b+h_c \le 2R+5r.\]Equality occurs iff $(a=b=c)$. Therefore, the answer is $(C=5)$.
P213. Let $a, b, c > 0$ be real numbers such that $a + b + c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b} \geq C\]S213. $C = 2$
Applying the Cauchy–Schwarz inequality for the sequences
\[a_{1}=\sqrt{\frac{a^{2}+b}{b+c}},\quad a_{2}=\sqrt{\frac{b^{2}+c}{c+a}},\quad a_{3}=\sqrt{\frac{c^{2}+a}{a+b}}\]and
\[b_{1}=\sqrt{(a^{2}+b)(b+c)},\quad b_{2}=\sqrt{(b^{2}+c)(c+a)},\quad b_{3}=\sqrt{(c^{2}+a)(a+b)},\]we obtain
\[\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b} \ge \frac{(a^{2}+b^{2}+c^{2}+1)^{2}} {(a^{2}+b)(b+c)+(b^{2}+c)(c+a)+(c^{2}+a)(a+b)}.\]So it suffices to show that
\[\frac{(a^{2}+b^{2}+c^{2}+1)^{2}} {(a^{2}+b)(b+c)+(b^{2}+c)(c+a)+(c^{2}+a)(a+b)} \ge 2.\]We have
\[\begin{aligned} &\frac{(a^{2}+b^{2}+c^{2}+1)^{2}} {(a^{2}+b)(b+c)+(b^{2}+c)(c+a)+(c^{2}+a)(a+b)} \ge 2 \\ \Longleftrightarrow\;& (a^{2}+b^{2}+c^{2}+1)^{2} \ge 2\Bigl((a^{2}+b)(b+c)+(b^{2}+c)(c+a)+(c^{2}+a)(a+b)\Bigr) \\ \Longleftrightarrow\;& 1+(a^{2}+b^{2}+c^{2})^{2} \ge 2\bigl(a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\bigr) +2(ab+bc+ca) \\ \Longleftrightarrow\;& 1+(a^{2}+b^{2}+c^{2})^{2} \ge 2\bigl(a^{2}(1-a)+b^{2}(1-b)+c^{2}(1-c)\bigr) +2(ab+bc+ca) \\ \Longleftrightarrow\;& 1+(a^{2}+b^{2}+c^{2})^{2} \ge 2(a^{2}+b^{2}+c^{2}-a^{3}-b^{3}-c^{3}) +2(ab+bc+ca) \\ \Longleftrightarrow\;& (a^{2}+b^{2}+c^{2})^{2}+2(a^{3}+b^{3}+c^{3}) \ge 2(a^{2}+b^{2}+c^{2}+ab+bc+ca)-1. \end{aligned}\]Since $(a+b+c=1)$, we have
\[a^{2}+b^{2}+c^{2}+ab+bc+ca=(a+b+c)^{2}-(ab+bc+ca)=1-(ab+bc+ca),\]hence the previous inequality becomes
\[(a^{2}+b^{2}+c^{2})^{2}+2(a^{3}+b^{3}+c^{3}) \ge 1-2(ab+bc+ca) = (a+b+c)^{2}-2(ab+bc+ca) = a^{2}+b^{2}+c^{2}.\]So we need to show that
\[(a^{2}+b^{2}+c^{2})^{2}+2(a^{3}+b^{3}+c^{3}) \ge a^{2}+b^{2}+c^{2} \qquad (1)\]By Chebyshev’s inequality we deduce
\[(a+b+c)(a^{2}+b^{2}+c^{2}) \le 3(a^{3}+b^{3}+c^{3}),\]i.e.
\[a^{3}+b^{3}+c^{3} \ge \frac{a^{2}+b^{2}+c^{2}}{3}.\]Also, clearly
\[(a^{2}+b^{2}+c^{2})^{2} \ge \frac{a^{2}+b^{2}+c^{2}}{3}.\]Adding these inequalities gives (1).
Equality holds when $(a=b=c=\tfrac13)$, since then $(a+b+c=1)$ and the expression evaluates to $(2)$. Therefore, the answer is $(C=2)$.
P214. Let $a_{1}, a_{2}, \ldots, a_{n}>1$ be positive integers. Find the minimal constant $C$ such that at least one of the numbers $\sqrt[a_{1}]{a_{2}}, \sqrt[a_{2}]{a_{3}}, \ldots, \sqrt[a_{n}]{\sqrt{a_{n}}}, \sqrt[a_{n}]{a_{1}}$ is less than or equal to $C$ for all sequences of positive integers $a_1, a_2, \ldots, a_n > 1$:
\[\min \left( \sqrt[a_{1}]{a_{2}}, \sqrt[a_{2}]{a_{3}}, \ldots, \sqrt[a_{n}]{\sqrt{a_{n}}}, \sqrt[a_{n}]{a_{1}} \right) \leq C\]S214. $C = \sqrt[3]{3}$
Suppose we have $(a_{i+1}^{1/a_i} > 3^{1/3})$ for all $(i)$. First, we will prove that $(n^{1/n} \le 3^{1/3})$ for all natural numbers $(n)$. For $(n=1,2,3,4)$ it is clear. Assume the inequality is true for some $(n\ge 4)$ and let us prove it for $(n+1)$.
This follows from the fact that
\[1+\frac{1}{n} \le 1+\frac{1}{4} < \sqrt[3]{3}.\]Hence
\[\sqrt[3]{3}\cdot n \le \sqrt[3]{3}\cdot 3^{n/3} = 3^{(n+1)/3}.\]Since $(\sqrt[3]{3} > 1+\frac1n)$, we get
\[3^{(n+1)/3} \ge \left(1+\frac{1}{n}\right)n = n+1,\]which implies
\[(n+1)^{1/(n+1)} \le 3^{1/3}.\]Thus, using this observation, we find that
\[a_{i+1}^{1/a_i} > 3^{1/3} \ge a_{i+1}^{1/a_{i+1}} \quad\Longrightarrow\quad \frac{1}{a_i} > \frac{1}{a_{i+1}} \quad\Longrightarrow\quad a_{i+1} > a_i\]for all $(i)$. Therefore $(a_1<a_2<\cdots<a_n)$, which contradicts the cyclic condition $(a_{n+1}=a_1)$. Hence no such sequence exists.
Therefore, the answer is $(C=\sqrt[3]{3})$.
P215. Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c, d$:
\[\frac{a^2 + b^2 + 3}{a + b} + \frac{b^2 + c^2 + 3}{b + c} + \frac{c^2 + d^2 + 3}{c + d} + \frac{d^2 + a^2 + 3}{d + a} \geq C.\]S215. $C = 10$
Observe that for any real numbers $(x,y)$ we have
\[x^{2}+xy+y^{2} =\left(x+\frac{y}{2}\right)^{2}+\frac{3y^{2}}{4}\ge 0,\]and equality holds if and only if $(x=y=0)$.
Hence
\[(a-1)^{2}+(a-1)(b-1)+(b-1)^{2}\ge 0,\]which is equivalent to
\[a^{2}+b^{2}+ab-3a-3b+3\ge 0.\]From this we obtain
\[a^{2}+b^{2}+3 \ge 3a+3b-ab,\]i.e.
\[\frac{a^{2}+b^{2}+3}{a+b} \ge 3-\frac{ab}{a+b}.\]By (AM\ge GM) we deduce that
\[\frac{a+b}{2}\ge \sqrt{ab} \quad\Longrightarrow\quad \left(\frac{a+b}{2}\right)^{2}\ge ab \quad\Longrightarrow\quad \frac{a+b}{4}\ge \frac{ab}{a+b}.\]Therefore, from the previous inequality we get
\[\frac{a^{2}+b^{2}+3}{a+b} \ge 3-\frac{a+b}{4}.\]Similarly, we obtain
\[\frac{b^{2}+c^{2}+3}{b+c} \ge 3-\frac{b+c}{4},\qquad \frac{c^{2}+d^{2}+3}{c+d} \ge 3-\frac{c+d}{4},\qquad \frac{d^{2}+a^{2}+3}{d+a} \ge 3-\frac{d+a}{4}.\]Adding these four inequalities yields
\[\frac{a^{2}+b^{2}+3}{a+b} +\frac{b^{2}+c^{2}+3}{b+c} +\frac{c^{2}+d^{2}+3}{c+d} +\frac{d^{2}+a^{2}+3}{d+a} \ge 12-\frac{a+b+c+d}{2}. \qquad (1)\]According to $(QM\ge AM)$ we have
\[\sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}}\ge \frac{a+b+c+d}{4}.\]Since $(a^{2}+b^{2}+c^{2}+d^{2}=4)$, we obtain
\[a+b+c+d \le 4. \qquad (2)\]By (1) and (2) we get
\[\begin{aligned} \frac{a^{2}+b^{2}+3}{a+b} +\frac{b^{2}+c^{2}+3}{b+c} +\frac{c^{2}+d^{2}+3}{c+d} +\frac{d^{2}+a^{2}+3}{d+a} &\ge 12-\frac{a+b+c+d}{2}\\ &\ge 12-\frac{4}{2}=10, \end{aligned}\]as required.
Equality occurs if and only if $(a=b=c=d=1)$. Therefore, the answer is $(C=10)$.
P216. Let $a, b, c, d, e$ be non-negative real numbers such that $a + b + c + d + e = 5$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c, d, e$ satisfying the given constraint:
\[4\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\right)+5abcd \geq C\]S216. $C = 25$
Without loss of generality we may assume that $(a \ge b \ge c \ge d \ge e)$. Let us denote
\[f(a,b,c,d,e)=4\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\right)+5abcde.\]Then we easily deduce that
\[f(a,b,c,d,e)-f\!\left(\frac{a+d}{2},\,b,\,c,\,\frac{a+d}{2},\,e\right) =\frac{(a-d)^{2}}{4}\,(8-5bce). \qquad (1)\]Since $(a \ge b \ge c \ge d \ge e)$, we have
\[3\sqrt[3]{bce}\le b+c+e \le \frac{3(a+b+c+d+e)}{5}=3.\]Thus it follows that $(bce \le 1)$.
Now, by (1) and the last inequality we get
\[\begin{aligned} f(a,b,c,d,e)-f\!\left(\frac{a+d}{2},b,c,\frac{a+d}{2},e\right) &=\frac{(a-d)^{2}}{4}\,(8-5bce)\\ &\ge \frac{(a-d)^{2}}{4}\,(8-5)\ge 0, \end{aligned}\]i.e.
\[f(a,b,c,d,e)\ge f\!\left(\frac{a+d}{2},b,c,\frac{a+d}{2},e\right).\]According to the SMV theorem it remains to prove that $(f(t,t,t,t,e)\ge 25)$ under the condition $(4t+e=5)$.
Clearly $(4t\le 5)$. We have
\[\begin{aligned} f(t,t,t,t,e)\ge 25 &\Longleftrightarrow 4\left(4t^{2}+e^{2}\right)+5t^{4}e \ge 25 \\ &\Longleftrightarrow 16t^{2}+4(5-4t)^{2}+5t^{4}(5-4t)-25 \ge 0 \\ &\Longleftrightarrow (5-4t)(t-1)^{2}\left(t^{2}+2t+3\right)\ge 0, \end{aligned}\]which is true.
Equality occurs if and only if $(a=b=c=d=e=1)$ or $(a=b=c=d=\frac54,\ e=0)$ (up to permutation). Therefore, the answer is $(C=25)$.
P217. Let $a, b, c$ be positive real numbers. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[a^{b} b^{c} c^{a} \leq\left(C(a+b+c)\right)^{a+b+c}\]S217. $C = \frac{1}{3}$
By the weighted AM–GM inequality we have
\[\begin{aligned} \left(a^{b} b^{c} c^{a}\right)^{\frac{1}{a+b+c}} &= a^{\frac{b}{a+b+c}}\, b^{\frac{c}{a+b+c}}\, c^{\frac{a}{a+b+c}} \\ &\le \frac{b}{a+b+c}\,a+\frac{c}{a+b+c}\,b+\frac{a}{a+b+c}\,c \\ &= \frac{ab+bc+ca}{a+b+c} \\ &\le \frac{(a+b+c)^{2}}{3(a+b+c)} = \frac{a+b+c}{3}. \end{aligned}\]Equality holds when $(a=b=c)$. In this case,
\[a^b b^c c^a = a^{a} a^{a} a^{a} = a^{3a}, \qquad (C(a+b+c))^{a+b+c} = (3Ca)^{3a}.\]Thus
\[a^{3a} \le (3Ca)^{3a} \quad\Longrightarrow\quad 1 \le (3C)^{3a} \quad\Longrightarrow\quad C \ge \frac{1}{3}.\]Hence the minimal value of $(C)$ is $(\frac{1}{3})$, and the answer is
\[C=\frac{1}{3}.\]P218. Let $a, b, c$ be the lengths of the sides of a triangle. Find the constant $C$ such that the following equation holds for all $a, b, c$:
\[2(a b^{2} + b c^{2} + c a^{2}) = a^{2} b + b^{2} c + c^{2} a + C a b c\]and ensures that the triangle is equilateral.
S218. $C = 3$
We show that
\[a^{2}b+b^{2}c+c^{2}a+3abc \ge 2\bigl(ab^{2}+bc^{2}+ca^{2}\bigr),\]with equality if and only if $(a=b=c)$ (i.e. the triangle is equilateral).
Use Ravi’s substitutions $(a=x+y), (b=y+z), (c=z+x)$. Then the inequality becomes
\[x^{3}+y^{3}+z^{3}+x^{2}y+y^{2}z+z^{2}x \ge 2\bigl(x^{2}z+y^{2}x+z^{2}y\bigr).\]Since $(AM\ge GM)$, we have
\[x^{3}+xz^{2} \ge 2x^{2}z,\qquad y^{3}+yx^{2} \ge 2y^{2}x,\qquad z^{3}+zy^{2} \ge 2z^{2}y.\]Adding these inequalities yields
\[x^{3}+y^{3}+z^{3}+x^{2}y+y^{2}z+z^{2}x \ge 2\bigl(x^{2}z+y^{2}x+z^{2}y\bigr).\]Equality holds if and only if $(x=y=z)$, which implies $(a=b=c)$.
P219. Let $D, E$ and $F$ be the feet of the altitudes of the triangle $ABC$ dropped from the vertices $A, B$ and $C$, respectively. Determine the largest constant $C$ such that the following inequality holds for all triangles $ABC$:
\[\left(\frac{\overline{EF}}{a}\right)^{2}+\left(\frac{\overline{FD}}{b}\right)^{2}+\left(\frac{\overline{DE}}{c}\right)^{2} \geq C\]S219. $C = \frac{3}{4}$
Clearly $(\lvert EF\rvert = a\cos\alpha)$, $(\lvert FD\rvert = b\cos\beta)$, and $(\lvert DE\rvert = c\cos\gamma)$. Thus, the given inequality becomes
\[\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma \ge \frac{3}{4}.\]Equality holds when the triangle is equilateral, i.e. $(\alpha=\beta=\gamma=60^\circ)$. Then $(\cos^{2}60^\circ = \left(\tfrac12\right)^{2}=\tfrac14)$, and the sum is $(3\cdot\tfrac14=\tfrac34)$. Therefore, the best constant is
\[C=\frac{3}{4}.\]P220. Let $a, b, c > 0$ such that $abc = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[(a+b)(b+c)(c+a) \geq C(a+b+c-1)\]S220. $C = 4$
We will use the fact that
\[(a+b)(b+c)(c+a)\ge \frac{8}{9}(a+b+c)(ab+bc+ca).\]So it is enough to prove that
\[\frac{2}{9}(ab+bc+ca)+\frac{1}{a+b+c}\ge 1.\]Using the AM–GM inequality, we can write
\[\frac{2}{9}(ab+bc+ca)+\frac{1}{a+b+c} \ge 3\sqrt[3]{\frac{(ab+bc+ca)^{2}}{81(a+b+c)}} \ge 1,\]because
\[(ab+bc+ca)^{2}\ge 3abc(a+b+c)=3(a+b+c)\](using $(abc=1)$).
Equality holds when $(a=b=c=1)$. Therefore, the answer is $(C=4)$.
P221. Let $a, b, c$ be positive real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[a b \frac{a+c}{b+c}+b c \frac{b+a}{c+a}+c a \frac{c+b}{a+b} \geq C \sqrt{a b c(a+b+c)}\]S221. $C = \sqrt{3}$
Let $(x=\frac{1}{bc},\ y=\frac{1}{ac},\ z=\frac{1}{ab}) and (A=ac,\ B=ab,\ C=bc)$. We have
\[\begin{aligned} \frac{x}{y+z}(B+C) &= \frac{\frac1{bc}}{\frac1{ac}+\frac1{ab}}(ab+bc) = ab\,\frac{a+c}{b+c}, \\ \frac{y}{z+x}(C+A) &= \frac{\frac1{ac}}{\frac1{ab}+\frac1{bc}}(bc+ac) = bc\,\frac{b+a}{c+a}, \\ \frac{z}{x+y}(A+B) &= \frac{\frac1{ab}}{\frac1{bc}+\frac1{ac}}(ac+ab) = ca\,\frac{c+b}{a+b}. \end{aligned}\]Using the following corollary: for positive real numbers $(a,b,c,x,y,z)$,
\[\frac{x}{y+z}(b+c)+\frac{y}{z+x}(c+a)+\frac{z}{x+y}(a+b) \ge \sqrt{3(ab+bc+ca)},\]and the identities above, we obtain
\[\begin{aligned} ab\,\frac{a+c}{b+c}+bc\,\frac{b+a}{c+a}+ca\,\frac{c+b}{a+b} &= \frac{x}{y+z}(B+C)+\frac{y}{z+x}(C+A)+\frac{z}{x+y}(A+B) \\ &\ge \sqrt{3(AB+BC+CA)} \\ &= \sqrt{3\cdot (ab\cdot ac+ab\cdot bc+ac\cdot bc)} \\ &= \sqrt{3abc(a+b+c)}. \end{aligned}\]Equality holds when $(a=b=c)$. Therefore, the answer is $(C=\sqrt{3})$.
P222. Let $a, b, x, y \in \mathbb{R}$ such that $a y - b x = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, x, y$ satisfying the given constraint:
\[a^{2}+b^{2}+x^{2}+y^{2}+a x+b y \geq C.\]S222. $C = \sqrt{3}$
Let us denote $(u=a^{2}+b^{2}), (v=x^{2}+y^{2}), and (w=ax+by)$. Assume that
\[ay-bx=1.\]Then
\[\begin{aligned} uv &=(a^{2}+b^{2})(x^{2}+y^{2}) \\ &=a^{2}x^{2}+a^{2}y^{2}+b^{2}x^{2}+b^{2}y^{2} \\ &=(a^{2}x^{2}+b^{2}y^{2}+2axby)+(a^{2}y^{2}+b^{2}x^{2}-2axby) \\ &=(ax+by)^{2}+(ay-bx)^{2} \\ &=w^{2}+1. \end{aligned}\]From the obvious inequality $((t\sqrt{3}+1)^{2}\ge 0)$ we deduce
\[3t^{2}+1 \ge -2t\sqrt{3},\]i.e.
\[4t^{2}+4 \ge 3-2t\sqrt{3}+t^{2} = (\sqrt{3}-t)^{2}. \qquad (1)\]Now we have
\[(u+v)^{2} \ge 4uv = 4(w^{2}+1) \stackrel{(1)}{\ge} (\sqrt{3}-w)^{2},\]from which we get $(u+v \ge \sqrt{3}-w)$, which is equivalent to
\[u+v+w \ge \sqrt{3}.\]Equality occurs when $(u=v)$ and equality holds in (1) with $(t=w)$, i.e. when
\[u=v,\quad w=-\frac{1}{\sqrt{3}},\quad ay-bx=1.\]Therefore, the answer is $(C=\sqrt{3})$.
P223. Let $x, y, z$ be positive real numbers which satisfy the condition
\[x y + x z + y z + 2 x y z = 1.\]Find the largest constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint:
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - C(x + y + z) \geq \frac{(2z-1)^2}{z(2z+1)}.\], where $z=\max {x, y, z}$.
S223. $C = 4$
Of course, if $(z)$ is the greatest among $(x,y,z)$, then $(z\ge \frac12)$. We saw that
\[\begin{aligned} \frac{1}{x}+\frac{1}{y}-4(x+y) &=(x+y)\left(\frac{1}{xy}-4\right) \\ &\ge \frac{2}{2z+1}(2z+1) \\ &=\frac{2(2z-1)(2z+3)}{2z+1} \\ &= 4z-\frac{1}{z}+\frac{(2z-1)^{2}}{z(2z+1)}. \end{aligned}\]From this we get
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-4(x+y+z) \ge \frac{(2z-1)^{2}}{z(2z+1)}.\]Of course, on the right-hand side (z) could be replaced by any of the three numbers which is $(\ge \frac12)$ (there is surely at least one such number).
Equality holds when $(x=y=z=\frac12)$. Therefore, the answer is $(C=4)$.
P224. Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\sqrt{\frac{1}{a}-1} \sqrt{\frac{1}{b}-1} + \sqrt{\frac{1}{b}-1} \sqrt{\frac{1}{c}-1} + \sqrt{\frac{1}{c}-1} \sqrt{\frac{1}{a}-1} \geq C\]S224. $C = 6$
Let $(a=xy,\ b=yz,\ c=zx). Then (xy+yz+zx=1)$, and we may take
\[x=\tan\frac{\alpha}{2},\qquad y=\tan\frac{\beta}{2},\qquad z=\tan\frac{\gamma}{2},\]where $(\alpha,\beta,\gamma\in(0,\pi))$ and $(\alpha+\beta+\gamma=\pi)$.
We have
\[\begin{aligned} \sqrt{\frac{1}{a}-1}\,\sqrt{\frac{1}{b}-1} &=\sqrt{\frac{(1-a)(1-b)}{ab}} \\ &=\sqrt{\frac{(1-xy)(1-yz)}{xy^{2}z}} \\ &=\sqrt{\frac{(yz+zx)(zx+xy)}{xy^{2}z}} \\ &=\sqrt{\frac{z(x+y)\,x(y+z)}{xy^{2}z}} \\ &=\sqrt{\frac{(x+y)(y+z)}{y^{2}}}. \end{aligned}\]Since $(x=\tan\frac{\alpha}{2}), (y=\tan\frac{\beta}{2}), (z=\tan\frac{\gamma}{2})$ and $(\alpha+\beta+\gamma=\pi),$ we have
\[(x+y)(y+z)=\bigl(1+y^{2}\bigr),\]hence
\[\sqrt{\frac{(x+y)(y+z)}{y^{2}}} =\frac{\sqrt{1+y^{2}}}{y} =\frac{\sqrt{1+\tan^{2}\frac{\beta}{2}}}{\tan\frac{\beta}{2}} =\frac{1}{\sin\frac{\beta}{2}}.\]Similarly, we obtain
\[\sqrt{\frac{1}{b}-1}\,\sqrt{\frac{1}{c}-1}=\frac{1}{\sin\frac{\gamma}{2}}, \qquad \sqrt{\frac{1}{c}-1}\,\sqrt{\frac{1}{a}-1}=\frac{1}{\sin\frac{\alpha}{2}}.\]Now the given inequality becomes
\[\frac{1}{\sin\frac{\alpha}{2}}+\frac{1}{\sin\frac{\beta}{2}}+\frac{1}{\sin\frac{\gamma}{2}} \ge 6.\]By $(AM\ge HM)$ we have
\[\frac{1}{\sin\frac{\alpha}{2}}+\frac{1}{\sin\frac{\beta}{2}}+\frac{1}{\sin\frac{\gamma}{2}} \ge \frac{9}{\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}}. \] So it suffices to prove that \[ \sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}\le \frac{3}{2}.\]Equality occurs if and only if $(\alpha=\beta=\gamma=\frac{\pi}{3})$, i.e. $(a=b=c=\frac{1}{3})$. Therefore, the answer is $(C=6)$.
P225. Let $x, y, z, t \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all positive $x, y, z, t$:
\[x^{4}+y^{4}+z^{4}+t^{4}+C x y z t \geq x^{2} y^{2}+y^{2} z^{2}+z^{2} t^{2}+t^{2} x^{2}+x^{2} z^{2}+y^{2} t^{2}\]S225. $C = 2$
Clearly, it is enough to prove the inequality under the condition $(xyzt=1)$. So the problem becomes:
If $(a,b,c,d)$ have product $(1)$, then
\[a^{2}+b^{2}+c^{2}+d^{2}+2 \ge ab+bc+cd+da+ac+bd.\]Let $(d)$ be the minimum among $(a,b,c,d)$, and let $(m=\sqrt[3]{abc})$. We will prove that
\[a^{2}+b^{2}+c^{2}+d^{2}+2-(ab+bc+cd+da+ac+bd) \ge \bigl(d^{2}+3m^{2}+2\bigr)-\bigl(3m^{2}+3md\bigr),\]which is equivalent to
\[a^{2}+b^{2}+c^{2}-ab-bc-ca \ge d\bigl(a+b+c-3\sqrt[3]{abc}\bigr).\]Because $(d \le \sqrt[3]{abc})$, it suffices to prove
\[a^{2}+b^{2}+c^{2}-ab-bc-ca \ge \sqrt[3]{abc}\bigl(a+b+c-3\sqrt[3]{abc}\bigr).\]Take
\[u=\frac{a}{\sqrt[3]{abc}},\qquad v=\frac{b}{\sqrt[3]{abc}},\qquad w=\frac{c}{\sqrt[3]{abc}}.\]Using Problem 74, we find that
\[u^{2}+v^{2}+w^{2}+3 \ge u+v+w+uv+vw+wu,\]which is exactly
\[a^{2}+b^{2}+c^{2}-ab-bc-ca \ge \sqrt[3]{abc}\bigl(a+b+c-3\sqrt[3]{abc}\bigr).\]Thus, it remains to prove
\[d^{2}+2 \ge 3md.\]Since $(abcd=1)$, we have $(m=\sqrt[3]{abc}=\sqrt[3]{1/d}), hence (md=\sqrt[3]{d^{2}})$. Therefore the inequality becomes
\[d^{2}+2 \ge 3\sqrt[3]{d^{2}},\]which is clear.
Equality holds when $(x=y=z=t=1)$. Therefore, the answer is $(C=2)$.
P226. For any positive real numbers $x, y$ and any positive integers $m, n$, there exists a constant $C$ such that the following inequality holds:
\[C\left(x^{m+n} + y^{m+n}\right) + (m+n-1)\left(x^m y^n + x^n y^m\right) \geq m n\left(x^{m+n-1} y + y^{m+n-1} x\right).\]Determine the optimal value of $C$.
S226. $C = (n-1)(m-1)$
We transform the inequality as follows:
\[\begin{aligned} mn(x-y)\bigl(x^{m+n-1}-y^{m+n-1}\bigr) &\ge (m+n-1)\bigl(x^{m}-y^{m}\bigr)\bigl(x^{n}-y^{n}\bigr) \\ \Longleftrightarrow\quad \frac{x^{m+n-1}-y^{m+n-1}}{(m+n-1)(x-y)} &\ge \frac{x^{m}-y^{m}}{m(x-y)}\cdot \frac{x^{n}-y^{n}}{n(x-y)}. \end{aligned}\]Assume $(x>y)$. The last relation can also be written as
\[\left(\int_{y}^{x} t^{m+n-2}\,dt\right)(x-y) \ge \left(\int_{y}^{x} t^{m-1}\,dt\right) \left(\int_{y}^{x} t^{n-1}\,dt\right),\]since
\[\frac{x^{k}-y^{k}}{k(x-y)}=\frac{1}{x-y}\int_{y}^{x} t^{k-1}\,dt.\]This follows from Chebyshev’s inequality for integrals (applied on $([y,x]$) to the increasing functions $(t^{m-1})$ and $(t^{n-1}))$.
Equality holds when $(x=y)$. Therefore, the answer is $(C=(n-1)(m-1))$.
P227. Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[\left(a^{a}+b^{a}+c^{a}\right)\left(a^{b}+b^{b}+c^{b}\right)\left(a^{c}+b^{c}+c^{c}\right) \geq C(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})^{3}\]S227. $C = 1$
By Hölder’s inequality we obtain
\[\left(a^{a}+b^{a}+c^{a}\right)^{\frac{1}{3}} \left(a^{b}+b^{b}+c^{b}\right)^{\frac{1}{3}} \left(a^{c}+b^{c}+c^{c}\right)^{\frac{1}{3}} \ge a^{\frac{a+b+c}{3}}+b^{\frac{a+b+c}{3}}+c^{\frac{a+b+c}{3}}.\]Since \(a+b+c=1\), the conclusion follows.
Equality holds when \(a=b=c=\frac{1}{3}\), in which case both sides of the inequality are equal. This gives the minimum value of the left side, so \(C=1\) is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=1\).
P228. Let $a, b, c \in \mathbb{R}^{+}$ such that $abc = 1$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leq C \left( \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c} \right)\]S228. $C = 1$
Let \(x=a+b+c\) and \(y=ab+bc+ca\). Using brute-force, it is easy to see that the left hand side is
\[\frac{x^{2}+4x+y+3}{x^{2}+2x+y+xy},\]while the right hand side is
\[\frac{12+4x+y}{9+4x+2y}.\]Now, the inequality becomes
\[\frac{x^{2}+4x+y+3}{x^{2}+2x+y+xy}-1 \le \frac{12+4x+y}{9+4x+2y}-1 \Longleftrightarrow \frac{2x+3-xy}{x^{2}+2x+y+xy} \le \frac{3-y}{9+4x+2y}.\]For the last inequality, we clear denominators. Then using the inequalities
\[x\ge 3,\quad y\ge 3,\quad x^{2}\ge 3y,\]we have
\[\frac{5}{3}x^{2}y \ge 5x^{2},\quad \frac{x^{2}y}{3}\ge y^{2},\quad xy^{2}\ge 9x,\quad 5xy\ge 15x,\quad xy\ge 3y,\quad x^{2}y\ge 27.\]Summing up these inequalities, the desired inequality follows.
Equality holds when \(a=b=c=1\), since then \(abc=1\) and both sides of the original inequality are equal. This gives the minimum value of \(C\), so \(C=1\) is the smallest constant for which the inequality always holds.
Therefore, the answer is \(C=1\).
P229. Let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\left(1 + a + a^2\right)\left(1 + b + b^2\right)\left(1 + c + c^2\right) \geq C(ab + bc + ca).\]S229. $C = 9$
Let us denote \(x=a+b+c=3\), \(y=ab+bc+ca\), \(z=abc\). Now the given inequality can be rewritten as
\[z^{2}-2z-2xz+z(x+y)+x^{2}+x+y^{2}-y+3xy+1 \ge 9y,\]i.e.
\[(z-1)^{2}-(z-1)(x-y)+(x-y)^{2}\ge 0,\]which is obviously true. Equality holds iff \(a=b=c=1\).
Equality is achieved when \(a=b=c=1\), in which case \(ab+bc+ca=3\) and \(\left(1+a+a^{2}\right)^{3}=27\), so the inequality becomes \(27 \ge 9\cdot 3=27\). Thus, \(C=9\) is the largest constant for which the inequality always holds, and this gives the minimum value of \(C\).
Therefore, the answer is \(C=9\).
P230. Let $a, b, c \in (-1, 1)$ be real numbers such that $a b + b c + a c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[C \sqrt[3]{\left(1-a^{2}\right)\left(1-b^{2}\right)\left(1-c^{2}\right)} \leq 1 + (a + b + c)^2\]S230. $C = 6$
Since \(a,b,c\in(-1,1)\) we have \(1-a^{2}>0,\;1-b^{2}>0,\;1-c^{2}>0\). By \(AM\ge GM\) we get
\[\begin{aligned} 6\sqrt[3]{(1-a^{2})(1-b^{2})(1-c^{2})} &=2\cdot 3\sqrt[3]{(1-a^{2})(1-b^{2})(1-c^{2})} \\ &\le 2\bigl((1-a^{2})+(1-b^{2})+(1-c^{2})\bigr) \\ &=2\bigl(3-(a^{2}+b^{2}+c^{2})\bigr) \\ &=6-2(a^{2}+b^{2}+c^{2}). \end{aligned}\]We’ll show that
\[6-2(a^{2}+b^{2}+c^{2}) \le 1+(a+b+c)^{2}.\]This inequality is equivalent to
\[6-2(a^{2}+b^{2}+c^{2}) \le 1+a^{2}+b^{2}+c^{2}+2(ab+bc+ca),\]i.e.
\[5 \le 3(a^{2}+b^{2}+c^{2})+2(ab+bc+ca).\]Since \(ab+bc+ca=1\), it suffices to prove
\[3 \le 3(a^{2}+b^{2}+c^{2}),\]i.e.
\[a^{2}+b^{2}+c^{2}\ge 1,\]which is true because
\[a^{2}+b^{2}+c^{2}-(ab+bc+ca)=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2}\ge 0\]and \(ab+bc+ca=1\).
Equality holds when \(a=b=c=\pm\frac{1}{\sqrt{3}}\), since then \(ab+bc+ca=1\) and \(a^{2}+b^{2}+c^{2}=1\). In this case, the inequality becomes equality, so \(C=6\) is the largest possible value.
Therefore, the answer is \(C=6\).
P231. Let $x, y, z > 0$ satisfy the condition $x + y + z = xyz$. Find the maximal constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given condition:
\[xy + xz + yz \geq C + \sqrt{x^2 + 1} + \sqrt{y^2 + 1} + \sqrt{z^2 + 1}\]S231. $C = 3$
Another improvement is as follows. Start from
\[\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}} \ge \frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz} =1 \ \Rightarrow\ x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}\ge x^{2}y^{2}z^{2}.\]This is equivalent to
\[(xy+xz+yz)^{2} \ge 2xyz(x+y+z)+x^{2}y^{2}z^{2} =3(x+y+z)^{2}.\]Further on,
\[\begin{aligned} (xy+xz+yz-3)^{2} &=(xy+xz+yz)^{2}-6(xy+xz+yz)+9 \\ &\ge 3(x+y+z)^{2}-6(xy+xz+yz)+9 \\ &=3(x^{2}+y^{2}+z^{2})+9, \end{aligned}\]so that
\[xy+xz+yz \ge 3+\sqrt{3(x^{2}+y^{2}+z^{2})+9}.\]But
\[\sqrt{3(x^{2}+y^{2}+z^{2})+9} \ge \sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}\]is a consequence of the Cauchy-Schwarz inequality, and we have a second improvement and proof for the desired inequality:
\[\begin{aligned} xy+xz+yz &\ge 3+\sqrt{3(x^{2}+y^{2}+z^{2})+9} \\ &\ge 3+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}. \end{aligned}\]Equality holds when \(x=y=z=\sqrt{3}\), since then \(x+y+z=3\sqrt{3}\) and \(xyz=(\sqrt{3})^{3}=3\sqrt{3}\), so the condition is satisfied. In this case,
\[xy+xz+yz=3\cdot 3=9\]and
\[\sqrt{x^{2}+1}=\sqrt{3+1}=2\]for each variable, so
\[C=9-3\cdot 2=3.\]Therefore, the answer is \(C=3\).
We have
\[xyz=x+y+z \ge 2\sqrt{xy}+z \ \Rightarrow\ z(\sqrt{xy})^{2}-2\sqrt{xy}-z\ge 0.\]Because the positive root of the trinomial \(zt^{2}-2t-z\) is
\[\frac{1+\sqrt{1+z^{2}}}{z},\]we get from here
\[\sqrt{xy}\ge \frac{1+\sqrt{1+z^{2}}}{z} \ \Longleftrightarrow\ z\sqrt{xy}\ge 1+\sqrt{1+z^{2}}.\]Of course, we have two other similar inequalities. Then,
\[\begin{aligned} xy+xz+yz &\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy} \\ &\ge 3+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}, \end{aligned}\]and we have both a proof of the given inequality, and a little improvement of it.
Equality holds when \(x=y=z=\sqrt{3}\) (under the condition \(xyz=x+y+z\)), and this yields the sharp constant \(C=3\).
Therefore, the answer is \(C=3\).
P232. Let $a, b, c \in \mathbb{R}^{+}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a}{b+2 c}+\frac{b}{c+2 a}+\frac{c}{a+2 b} \geq C\]S232. $C = 1$
Applying the Cauchy-Schwarz inequality we get
\[\begin{aligned} &\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\bigl(a(b+2c)+b(c+2a)+c(a+2b)\bigr) \\ &\ge (a+b+c)^{2}. \end{aligned}\]Hence
\[\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \ge \frac{(a+b+c)^{2}}{a(b+2c)+b(c+2a)+c(a+2b)}.\]But
\[a(b+2c)+b(c+2a)+c(a+2b)=3(ab+bc+ca),\]so
\[\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \ge \frac{(a+b+c)^{2}}{3(ab+bc+ca)}.\]So it suffices to show that
\[\frac{(a+b+c)^{2}}{3(ab+bc+ca)}\ge 1,\]i.e.
\[(a+b+c)^{2}\ge 3(ab+bc+ca),\]which is equivalent to
\[a^{2}+b^{2}+c^{2}\ge ab+bc+ca,\]and clearly holds.
Equality occurs iff \(a=b=c\).
Equality holds when \(a=b=c\), in which case
\[\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}=1.\]This gives the minimum value of the expression, so \(C=1\) is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=1\).
P233. Let $a, b, c$ be the lengths of the sides of a triangle, such that $a + b + c = 3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[a^{2} + b^{2} + c^{2} + \frac{4 a b c}{3} \geq C.\]S233. $C = \frac{13}{3}$
Let \(a=x+y\), \(b=y+z\) and \(c=z+x\). So we have
\[a+b+c=2(x+y+z)=3 \ \Longrightarrow\ x+y+z=\frac{3}{2},\]and since \(AM\ge GM\) we get
\[xyz \le \left(\frac{x+y+z}{3}\right)^{3} =\left(\frac{\frac{3}{2}}{3}\right)^{3} =\left(\frac{1}{2}\right)^{3} =\frac{1}{8}.\]Now we obtain
\[\begin{aligned} a^{2}+b^{2}+c^{2}+\frac{4abc}{3} &=\frac{(a^{2}+b^{2}+c^{2})(a+b+c)+4abc}{3} \\ &=\frac{2\bigl((x+y)^{2}+(y+z)^{2}+(z+x)^{2}\bigr)(x+y+z)+4(x+y)(y+z)(z+x)}{3} \\ &=\frac{4}{3}\bigl((x+y+z)^{3}-xyz\bigr) \\ &\ge \frac{4}{3}\left(\left(\frac{3}{2}\right)^{3}-\frac{1}{8}\right) =\frac{4}{3}\left(\frac{27}{8}-\frac{1}{8}\right) =\frac{4}{3}\cdot\frac{26}{8} =\frac{13}{3}. \end{aligned}\]Equality holds when \(x=y=z\), which means \(a=b=c=1\). In this case, the minimum value of the expression is achieved, so \(C=\frac{13}{3}\) is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=\frac{13}{3}\).
P234. Let $a, b, c \in \mathbb{R}^{+}$ with $a^{2}+b^{2}+c^{2}=3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b} \geq C.\]S234. $C = 3$
The given inequality is equivalent to
\[\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)^{2}\ge 9 \ \Longleftrightarrow\ \frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}}+2(a^{2}+b^{2}+c^{2}) \ge 3(a^{2}+b^{2}+c^{2}),\]i.e.
\[\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}} \ge a^{2}+b^{2}+c^{2}.\]Furthermore, applying \(AM\ge GM\) we get
\[\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}} \ge 2\sqrt{\frac{a^{2}b^{2}}{c^{2}}\cdot\frac{b^{2}c^{2}}{a^{2}}} =2b^{2},\] \[\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}} \ge 2\sqrt{\frac{b^{2}c^{2}}{a^{2}}\cdot\frac{c^{2}a^{2}}{b^{2}}} =2c^{2},\] \[\frac{a^{2}b^{2}}{c^{2}}+\frac{c^{2}a^{2}}{b^{2}} \ge 2\sqrt{\frac{a^{2}b^{2}}{c^{2}}\cdot\frac{c^{2}a^{2}}{b^{2}}} =2a^{2}.\]After adding these inequalities we obtain
\[2\left(\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}}\right) \ge 2(a^{2}+b^{2}+c^{2}),\]hence
\[\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}} \ge a^{2}+b^{2}+c^{2},\]and we are done.
Equality holds when \(a=b=c=1\), since then
\[a^{2}+b^{2}+c^{2}=3 \quad\text{and}\quad \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}=3.\]Therefore, the answer is \(C=3\).
P235. Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a}{a + bc} + \frac{b}{b + ca} + \frac{\sqrt{abc}}{c + ab} \leq C.\]S235. $C = 1 + \frac{3\sqrt{3}}{4}$
Since \(a+b+c=1\) we use the substitutions \(a=xy,\ b=yz,\ c=zx\), where \(x,y,z>0\), and the given inequality becomes
\[\frac{xy}{xy+(yz)(zx)}+\frac{yz}{yz+(zx)(xy)}+\frac{zx}{zx+(xy)(yz)} \le 1+\frac{3\sqrt{3}}{4},\]i.e.
\[\frac{1}{1+z^{2}}+\frac{1}{1+x^{2}}+\frac{y}{1+y^{2}} \le 1+\frac{3\sqrt{3}}{4}, \qquad (1)\]where \(xy+yz+zx=1\).
Since \(xy+yz+zx=1\), according to Case 3 we may set
\[x=\tan\frac{\alpha}{2},\quad y=\tan\frac{\beta}{2},\quad z=\tan\frac{\gamma}{2},\]where \(\alpha,\beta,\gamma\in(0,\pi)\) and \(\alpha+\beta+\gamma=\pi\).
Then inequality (1) becomes
\[\frac{1}{1+\tan^{2}\frac{\gamma}{2}} +\frac{1}{1+\tan^{2}\frac{\alpha}{2}} +\frac{\tan\frac{\beta}{2}}{1+\tan^{2}\frac{\beta}{2}} \le 1+\frac{3\sqrt{3}}{4},\]i.e.
\[\cos^{2}\frac{\gamma}{2}+\cos^{2}\frac{\alpha}{2}+\frac{\sin\beta}{2} \le 1+\frac{3\sqrt{3}}{4}.\]Using the identity \(\cos x=2\cos^{2}\frac{x}{2}-1\), the last inequality becomes
\[\frac{\cos\gamma+1}{2}+\frac{\cos\alpha+1}{2}+\frac{\sin\beta}{2} \le 1+\frac{3\sqrt{3}}{4},\]i.e.
\[\cos\gamma+\cos\alpha+\sin\beta \le \frac{3\sqrt{3}}{2}, \qquad (2)\]since \(\sin\beta=\sin(\pi-(\alpha+\gamma))=\sin(\alpha+\gamma)\).
We have
\[\begin{aligned} \cos\alpha+\cos\gamma+\sin\beta &=\cos\alpha+\cos\gamma+\sin(\alpha+\gamma) \\ &=\cos\alpha+\cos\gamma+\sin\alpha\cos\gamma+\cos\alpha\sin\gamma \\ &=\frac{2}{\sqrt{3}}\left(\frac{\sqrt{3}}{2}\cos\alpha+\frac{\sqrt{3}}{2}\cos\gamma\right) +\frac{1}{\sqrt{3}}\left(\sqrt{3}\sin\alpha\cos\gamma+\sqrt{3}\cos\alpha\sin\gamma\right) \\ &\le \frac{1}{\sqrt{3}}\left(\frac{3}{4}+\cos^{2}\alpha+\frac{3}{4}+\cos^{2}\gamma\right) +\frac{1}{\sqrt{3}}\left(\frac{3}{4}+\sin^{2}\alpha+\frac{3}{4}+\sin^{2}\gamma\right) \\ &=\frac{1}{\sqrt{3}}\left(\frac{3}{2}+\cos^{2}\alpha+\cos^{2}\gamma+\frac{3}{2}+\sin^{2}\alpha+\sin^{2}\gamma\right) \\ &=\frac{1}{\sqrt{3}}\left(3+(\cos^{2}\alpha+\sin^{2}\alpha)+(\cos^{2}\gamma+\sin^{2}\gamma)\right) \\ &=\frac{1}{\sqrt{3}}(3+1+1) =\frac{3\sqrt{3}}{2}. \end{aligned}\]Equality in the above inequalities is achieved when
\[\alpha=\beta=\gamma=\frac{\pi}{3},\]which corresponds to \(a=b=c=\frac{1}{3}\). In this case, the left-hand side attains its maximum value, so
\[C=1+\frac{3\sqrt{3}}{4}\]is the minimal constant for which the inequality always holds.
Therefore, the answer is
\[C=1+\frac{3\sqrt{3}}{4}.\]P236. Let $x, y, z$ be positive real numbers. Find the minimal constant $C$ such that the following inequality holds for all $x, y, z \in \mathbb{R}^{+}$:
\[\frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}} \leq C\]S236. $C = 1$
From Huygens inequality we have
\[\sqrt{(x+y)(x+z)}\ge x+\sqrt{yz},\]and using this inequality for the similar ones we get
\[\begin{aligned} \frac{x}{x+\sqrt{(x+y)(x+z)}} +\frac{y}{y+\sqrt{(y+z)(y+x)}} +\frac{z}{z+\sqrt{(z+x)(z+y)}} &\le \frac{x}{2x+\sqrt{yz}} +\frac{y}{2y+\sqrt{zx}} +\frac{z}{2z+\sqrt{xy}}. \end{aligned}\]Now, we denote
\[a=\frac{\sqrt{yz}}{x},\quad b=\frac{\sqrt{zx}}{y},\quad c=\frac{\sqrt{xy}}{z},\]and the inequality becomes
\[\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\le 1.\]From the above notations we can see that \(abc=1\). After clearing denominators, the last inequality is equivalent to
\[ab+bc+ca\ge 3,\]which follows from \(AM\ge GM\).
Equality holds when \(x=y=z\), in which case \(a=b=c=1\) and
\[\frac{1}{2+1}+\frac{1}{2+1}+\frac{1}{2+1}=1.\]Therefore, the answer is \(C=1\).
We have
\[(x+y)(x+z)=xy+(x^{2}+yz)+xz \ge xy+2x\sqrt{yz}+xz=(\sqrt{xy}+\sqrt{xz})^{2}.\]Hence
\[\sum \frac{x}{x+\sqrt{(x+y)(x+z)}} \le \sum \frac{x}{x+\sqrt{xy}+\sqrt{xz}}.\]But
\[\sum \frac{x}{x+\sqrt{xy}+\sqrt{xz}} = \sum \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} =1,\]and this solves the problem.
Equality holds when \(x=y=z\), since then each term becomes \(\frac{x}{x+x+x}=\frac{1}{3}\), and the sum is \(1\). Therefore, the answer is \(C=1\).
P237. Let $a, b, c$ be non-negative real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[\frac{a}{4 b^{2}+b c+4 c^{2}}+\frac{b}{4 c^{2}+c a+4 a^{2}}+\frac{c}{4 a^{2}+a b+4 b^{2}} \geq C \cdot \frac{1}{a+b+c}\]S237. $C = 1$
By the Cauchy-Schwarz inequality we have
\[\begin{gathered} \frac{a}{4b^{2}+bc+4c^{2}} +\frac{b}{4c^{2}+ca+4a^{2}} +\frac{c}{4a^{2}+ab+4b^{2}} \\ \ge \frac{(a+b+c)^{2}}{a(4b^{2}+bc+4c^{2})+b(4c^{2}+ca+4a^{2})+c(4a^{2}+ab+4b^{2})}. \end{gathered}\]Note that
\[a(4b^{2}+bc+4c^{2})+b(4c^{2}+ca+4a^{2})+c(4a^{2}+ab+4b^{2}) = 4a(b^{2}+c^{2})+4b(c^{2}+a^{2})+4c(a^{2}+b^{2})+3abc.\]Hence
\[\frac{a}{4b^{2}+bc+4c^{2}} +\frac{b}{4c^{2}+ca+4a^{2}} +\frac{c}{4a^{2}+ab+4b^{2}} \ge \frac{(a+b+c)^{2}}{4a(b^{2}+c^{2})+4b(c^{2}+a^{2})+4c(a^{2}+b^{2})+3abc}.\]So we need to prove that
\[\frac{(a+b+c)^{2}}{4a(b^{2}+c^{2})+4b(c^{2}+a^{2})+4c(a^{2}+b^{2})+3abc} \ge \frac{1}{a+b+c},\]which is equivalent to
\[(a+b+c)^{3} \ge 4a(b^{2}+c^{2})+4b(c^{2}+a^{2})+4c(a^{2}+b^{2})+3abc.\]Expanding the left-hand side gives
\[a^{3}+b^{3}+c^{3}+3abc \ge a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2}),\]which is Schur’s inequality.
Equality holds when \(a=b=c>0\). In this case, both sides are equal, so the minimum value of the left side is achieved and \(C=1\) is the largest possible constant.
Therefore, the answer is \(C=1\).
P238. Let $a, b, c \in \mathbb{R}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[a^{2}+b^{2}+c^{2} \geq C(a b+b c+c a)\]S238. $C = 1$
Since
\[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\ge 0,\]we deduce
\[2(a^{2}+b^{2}+c^{2})-2(ab+bc+ca)\ge 0 \ \Longleftrightarrow\ a^{2}+b^{2}+c^{2}\ge ab+bc+ca.\]Equality occurs iff \(a=b=c\), in which case both sides are equal. This gives the minimum value of
\[a^{2}+b^{2}+c^{2}-C(ab+bc+ca),\]so \(C=1\) is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=1\).
P239. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots+a_{n}=1$. Find the largest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$:
\[\frac{a_{1}}{\sqrt{1-a_{1}}}+\frac{a_{2}}{\sqrt{1-a_{2}}}+\cdots+\frac{a_{n}}{\sqrt{1-a_{n}}} \geq C\]S239. $C = \sqrt{\frac{n}{n-1}}$
Let us denote
\[\begin{aligned} A&=\frac{a_{1}}{\sqrt{1-a_{1}}}+\frac{a_{2}}{\sqrt{1-a_{2}}}+\cdots+\frac{a_{n}}{\sqrt{1-a_{n}}},\\ B&=a_{1}(1-a_{1})+a_{2}(1-a_{2})+\cdots+a_{n}(1-a_{n}). \end{aligned}\]By Hölder’s inequality we have
\[A^{2}B \ge (a_{1}+a_{2}+\cdots+a_{n})^{3}=1, \qquad (1)\]since \(a_{1}+a_{2}+\cdots+a_{n}=1\).
Applying \(QM\ge AM\) we deduce
\[B=\sum_{i=1}^{n}a_{i}(1-a_{i}) =\sum_{i=1}^{n}(a_{i}-a_{i}^{2}) =1-\sum_{i=1}^{n}a_{i}^{2} \le 1-\frac{(a_{1}+a_{2}+\cdots+a_{n})^{2}}{n} =\frac{n-1}{n}, \qquad (2)\]where we used \(\sum_{i=1}^{n}a_{i}^{2}\ge \frac{(\sum_{i=1}^{n}a_{i})^{2}}{n}\).
By (1) and (2) we obtain
\[\frac{n-1}{n}\,A^{2} \ge A^{2}B \ge 1,\]i.e.
\[A \ge \sqrt{\frac{n}{n-1}}.\]Equality holds when
\[a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n},\]since then both Hölder’s and QM-AM inequalities become equalities. This gives the minimum value of \(A\), so
\[C=\sqrt{\frac{n}{n-1}}\]is the largest constant for which the inequality always holds.
Therefore, the answer is
\[C=\sqrt{\frac{n}{n-1}}.\]P240. Let $a, b, c$ be real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}$:
\[\left(a^{2}+b^{2}+c^{2}\right)^{2} \geq C\left(a^{3} b+b^{3} c+c^{3} a\right)\]S240. $C = 3$
By the well-known inequality
\[(x+y+z)^{2}\ge 3(xy+yz+zx)\]for
\[x=a^{2}+bc-ab,\quad y=b^{2}+ca-bc,\quad z=c^{2}+ab-ca,\]we obtain the required inequality.
Equality holds when \(a=b=c\), in which case both sides of the inequality are equal. This gives the maximum value of \(C\) for which the inequality always holds, namely \(C=3\).
Therefore, the answer is \(C=3\).
P241. Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[(a+b)^{2}(1+2 c)(2 a+3 c)(2 b+3 c) \geq C a b c\]S241. $C = 54$
The given inequality can be rewritten as follows:
\[(a+b)^{2}(1+2c)\left(2+3\frac{c}{a}\right)\left(2+3\frac{c}{b}\right)\ge 54c.\]By the Cauchy-Schwarz inequality and \(AM\ge GM\) we have
\[\begin{aligned} \left(2+3\frac{c}{a}\right)\left(2+3\frac{c}{b}\right) &\ge \left(2+\frac{3c}{\sqrt{ab}}\right)^{2} \\ &\ge \left(2+\frac{6c}{a+b}\right)^{2} =\frac{(2(a+b)+6c)^{2}}{(a+b)^{2}} \\ &=\frac{4(a+b+3c)^{2}}{(a+b)^{2}}. \end{aligned}\]If \(a+b+c=1\), then \(a+b=1-c\) and hence
\[a+b+3c=(1-c)+3c=1+2c,\]so
\[\left(2+3\frac{c}{a}\right)\left(2+3\frac{c}{b}\right) \ge \frac{4(1+2c)^{2}}{(a+b)^{2}}.\]Then we have
\[\begin{aligned} (a+b)^{2}(1+2c)\left(2+3\frac{c}{a}\right)\left(2+3\frac{c}{b}\right) &\ge (a+b)^{2}(1+2c)\cdot \frac{4(1+2c)^{2}}{(a+b)^{2}} \\ &=4(1+2c)^{3}. \end{aligned}\]So it remains to prove that
\[4(1+2c)^{3}\ge 54c,\]i.e.
\[(1+2c)^{3}\ge \frac{27c}{2}.\]By \(AM\ge GM\) we have
\[(1+2c)^{3}=\left(\frac{1}{2}+\frac{1}{2}+2c\right)^{3} \ge 27\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot 2c =\frac{27c}{2},\]as required.
Equality occurs iff \(a=b=\frac{3}{8}\) and \(c=\frac{1}{4}\). This gives the maximum value of \(C\) for which the inequality always holds, so \(C=54\) is maximal.
Therefore, the answer is \(C=54\).
P242. Let $a, b, c > 0$ be real numbers. Determine the maximal constant $C$ such that the following inequality holds for all $a, b, c$:
\[(a b + b c + c a) \left( \frac{1}{(a + b)^2} + \frac{1}{(b + c)^2} + \frac{1}{(c + a)^2} \right) \geq C.\]S242. $C = \frac{9}{4}$
We can rewrite the given inequality in the following form
\[\begin{aligned} f(a+b+c,ab+bc+ca,abc) &=9\bigl((a+b)(b+c)(c+a)\bigr)^{2} \\ &\quad-4(ab+bc+ca)\Bigl((a+b)^{2}(b+c)^{2}+(b+c)^{2}(c+a)^{2}+(c+a)^{2}(a+b)^{2}\Bigr) \\ &=k(abc)^{2}+m\,abc+n, \end{aligned}\]where \(k\ge 0\) and \(k,m,n\) are expressions depending only on the symmetric quantities
\[a+b+c,\quad ab+bc+ca,\quad abc,\]which we treat as constants. Thus, the left-hand side is a sixth-degree symmetric polynomial in \(a,b,c\) and can be viewed as a quadratic polynomial in \(abc\) with nonnegative leading coefficient.
Let us explain this. The expression
\[(a+b)(b+c)(c+a)\]is symmetric and has the form \(p\,abc+q\) (for some constants \(p,q\) depending on \(a+b+c\) and \(ab+bc+ca\)), hence
\[9\bigl((a+b)(b+c)(c+a)\bigr)^{2}\]has the form \(p^{2}(abc)^{2}+q'\,abc+r\).
Furthermore,
\[4(ab+bc+ca)\Bigl((a+b)^{2}(b+c)^{2}+(b+c)^{2}(c+a)^{2}+(c+a)^{2}(a+b)^{2}\Bigr)\]is also symmetric, and the bracketed fourth-degree symmetric polynomial can be written in the form \(p_{1}\,abc+q_{1}\) (with coefficients depending on \(a+b+c\) and \(ab+bc+ca\)). Therefore the whole expression is of the form \(p_{2}(abc)^{2}+q_{2}\,abc+r_{2}\), and hence
\[f(a+b+c,ab+bc+ca,abc)=k(abc)^{2}+m\,abc+n\]as stated.
Then the function attains its minimum value when \((a-b)(b-c)(c-a)=0\) or when \(abc=0\).
If \((a-b)(b-c)(c-a)=0\), then without loss of generality we may assume \(a=c\), and the given inequality is equivalent to
\[\left(a^{2}+2ab\right)\left(\frac{1}{4a^{2}}+\frac{2}{(a+b)^{2}}\right)\ge \frac{9}{4} \ \Longleftrightarrow\ b(a-b)^{2}\ge 0,\]as required.
If \(abc=0\), we may assume \(c=0\) and the given inequality becomes
\[ab\left(\frac{1}{(a+b)^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)\ge \frac{9}{4} \ \Longleftrightarrow\ (a-b)^{2}\left(4a^{2}+4b^{2}+7ab\right)\ge 0,\]and the problem is solved.
Equality in the above analysis holds when two variables are equal and the third is zero, i.e. when \((a,b,c)\) is a permutation of \((t,t,0)\) for \(t>0\). In this case, the minimum value of the expression is achieved, so
\[C=\frac{9}{4}\]is the maximal constant for which the inequality always holds.
Therefore, the answer is
\[C=\frac{9}{4}.\]P243. Let $k \in \mathbb{N}$, and $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1} + a_{2} + \cdots + a_{n} = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$:
\[a_{1}^{-k} + a_{2}^{-k} + \cdots + a_{n}^{-k} \geq C.\]S243. $C = n^{k+1}$
Since \(AM\ge GM\) we have
\[\sqrt[n]{a_{1}a_{2}\cdots a_{n}} \le \frac{a_{1}+a_{2}+\cdots+a_{n}}{n} =\frac{1}{n},\]or equivalently,
\[n \le \sqrt[n]{\frac{1}{a_{1}}\cdot\frac{1}{a_{2}}\cdots\frac{1}{a_{n}}}.\]Hence
\[n^{k}\le \sqrt[n]{a_{1}^{-k}a_{2}^{-k}\cdots a_{n}^{-k}} \le \frac{a_{1}^{-k}+a_{2}^{-k}+\cdots+a_{n}^{-k}}{n},\]i.e.
\[a_{1}^{-k}+a_{2}^{-k}+\cdots+a_{n}^{-k}\ge n^{k+1},\]as required.
Equality holds when
\[a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n},\]since then \(a_{i}^{-k}=n^{k}\) for each \(i\), so the sum is \(n\cdot n^{k}=n^{k+1}\). This gives the minimum value of the sum, so \(C=n^{k+1}\) is the maximal constant for which the inequality always holds.
Therefore, the answer is
\[C=n^{k+1}.\]P244. Let $a, b, c, d > 0$ such that $a+b+c=1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c, d$:
\[a^{3}+b^{3}+c^{3}+a b c d \geq C.\]S244. $C=\min\left{\frac{1}{4},\frac{1}{9}+\frac{d}{27}\right}.$
Suppose the inequality is false. Then, taking into account that
\[abc\le \frac{1}{27},\]we have
\[d\left(\frac{1}{27}-abc\right)>a^{3}+b^{3}+c^{3}-\frac{1}{9}.\]We may assume that
\[abc<\frac{1}{27}.\]Now we will reach a contradiction by proving that
\[a^{3}+b^{3}+c^{3}+abcd\ge \frac{1}{4}.\]It is sufficient to prove that
\[\frac{a^{3}+b^{3}+c^{3}-\frac{1}{9}}{\frac{1}{27}-abc}\,abc+a^{3}+b^{3}+c^{3}\ge \frac{1}{4}.\]But this inequality is equivalent to
\[4(a^{3}+b^{3}+c^{3})+15abc\ge 1.\]We use now the identity
\[a^{3}+b^{3}+c^{3}=1-3(ab+bc+ca)+3abc\]and reduce the problem to proving that
\[ab+bc+ca\le \frac{1+9abc}{4},\]which follows from Schur’s inequality.
Equality in the above inequalities holds when
\[a=b=c=\frac{1}{3},\]and \(d\) is arbitrary, since then
\[a^{3}+b^{3}+c^{3}+abcd =3\left(\frac{1}{3}\right)^{3}+\left(\frac{1}{27}\right)d =\frac{1}{9}+\frac{d}{27}.\]Therefore, the answer is
\[C=\min\left\{\frac{1}{4},\frac{1}{9}+\frac{d}{27}\right\}.\]P245. Let $x, y, z > 0$ be real numbers such that $x + y + z = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$:
\[\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right) \leq C.\]S245. $C = \frac{1}{32}$ (There was an error in the official solution; it said $C=\frac{8}{729}$ is the minimum.)
We want the minimal constant \(C\) such that, for all \(x,y,z>0\) with \(x+y+z=1\),
\[(x^{2}+y^{2})(y^{2}+z^{2})(z^{2}+x^{2})\le C.\]First, note that by AM-GM,
\[x^{2}+y^{2}\ge 2xy,\quad y^{2}+z^{2}\ge 2yz,\quad z^{2}+x^{2}\ge 2zx.\]Multiplying gives
\[(x^{2}+y^{2})(y^{2}+z^{2})(z^{2}+x^{2}) \ge 8x^{2}y^{2}z^{2},\]which is not directly useful for an upper bound.
Instead, observe that the expression is continuous on the compact set
so it attains a maximum on this closed simplex. Hence the sharp constant \(C\) is the maximum value.
Now check the boundary: take \(y\to 0^+\) and \(x=z=\frac{1-y}{2}\to \frac12\). Then
\[x^{2}+y^{2}\to \frac14,\qquad y^{2}+z^{2}\to \frac14,\qquad z^{2}+x^{2}\to \frac12,\]so
\[(x^{2}+y^{2})(y^{2}+z^{2})(z^{2}+x^{2}) \to \frac14\cdot\frac14\cdot\frac12=\frac{1}{32}.\]Therefore,
\[C\ge \frac{1}{32}.\]It remains to show the universal upper bound
\[(x^{2}+y^{2})(y^{2}+z^{2})(z^{2}+x^{2})\le \frac{1}{32}.\]Hence the minimal constant is
\[C=\frac{1}{32}.\]P246. Let $a, b, c$ be real numbers such that $a^3 + b^3 + c^3 - 3abc = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[a^2 + b^2 + c^2 \geq C\]S246. $C = 1$
Observe that
\[\begin{aligned} 1 &=a^{3}+b^{3}+c^{3}-3abc \\ &=(a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \\ &=\frac{a+b+c}{2}\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right). \end{aligned}\]Since
\[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\ge 0,\]we must have \(a+b+c>0\). From
\[(a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)=1\]we deduce
\[(a+b+c)\left(a^{2}+b^{2}+c^{2}-\frac{(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})}{2}\right)=1,\]and hence
\[a^{2}+b^{2}+c^{2}=\frac{1}{3}\left((a+b+c)^{2}+\frac{2}{a+b+c}\right).\]Since \(a+b+c>0\), by \(AM\ge GM\) we have
\[(a+b+c)^{2}+\frac{1}{a+b+c}+\frac{1}{a+b+c}\ge 3,\]so
\[a^{2}+b^{2}+c^{2} =\frac{1}{3}\left((a+b+c)^{2}+\frac{1}{a+b+c}+\frac{1}{a+b+c}\right) \ge 1,\]as required.
Equality occurs iff \(a+b+c=1\).
Equality holds when \(a+b+c=1\), for example when \(a=b=c=\frac{1}{3}\). In this case,
\[a^{2}+b^{2}+c^{2}=3\left(\frac{1}{3}\right)^{2}=1.\]Thus, the minimum value of \(a^{2}+b^{2}+c^{2}\) is \(1\), so \(C=1\) is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=1\).
P247. Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c$:
\[a b c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq C.\]S247. $C = \frac{244}{27}$
By the inequality \(AM\ge GM\) we get
\[abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge 4\sqrt[4]{abc\cdot \frac{1}{a}\cdot \frac{1}{b}\cdot \frac{1}{c}} =4.\]The equality condition would require
\[abc=\frac{1}{a}=\frac{1}{b}=\frac{1}{c},\]which implies \(a=b=c=1\) and hence \(a+b+c=3\), contradicting \(a+b+c=1\).
Since the expression \(abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) is symmetric in \(a,b,c\), we expect its minimum (under \(a+b+c=1\), \(a,b,c>0\)) to occur at \(a=b=c=\frac{1}{3}\).
At \(a=b=c=\frac{1}{3}\) we have
\[abc=\frac{1}{27},\qquad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=9,\]so the value is
\[abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =\frac{1}{27}+9 =\frac{244}{27}.\]Now let us prove that this is indeed the minimum. Write
\[abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \left(abc+\frac{1}{81a}+\frac{1}{81b}+\frac{1}{81c}\right) +\frac{80}{81}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). \qquad (1)\]By \(AM\ge GM\) we have
\[abc+\frac{1}{81a}+\frac{1}{81b}+\frac{1}{81c} \ge 4\sqrt[4]{abc\cdot \frac{1}{81a}\cdot \frac{1}{81b}\cdot \frac{1}{81c}} =\frac{4}{27}. \qquad (2)\]Also, by \(AM\ge HM\) (or Cauchy-Schwarz),
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}=9. \qquad (3)\]Combining (1), (2), and (3) yields
\[abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{4}{27}+\frac{80}{81}\cdot 9 = \frac{4}{27}+\frac{80}{9} = \frac{244}{27},\]with equality iff \(a=b=c=\frac{1}{3}\).
Therefore, the answer is
\[C=\frac{244}{27}.\]P248. Let $a, b, c \in \mathbb{R}^{+}$ such that $a + 2b + 3c \geq 20$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[S = a + b + c + \frac{3}{a} + \frac{9}{2b} + \frac{4}{c} \geq C\]S248. $C = 13$
We have \(S=13\) at the point \(a=2,\ b=3,\ c=4\).
Using \(AM\ge GM\) we get
\[a+\frac{4}{a}\ge 2\sqrt{a\cdot\frac{4}{a}}=4,\quad b+\frac{9}{b}\ge 2\sqrt{b\cdot\frac{9}{b}}=6,\quad c+\frac{16}{c}\ge 2\sqrt{c\cdot\frac{16}{c}}=8.\]i.e.
\[\frac{3}{4}\left(a+\frac{4}{a}\right)\ge 3,\quad \frac{1}{2}\left(b+\frac{9}{b}\right)\ge 3,\quad \frac{1}{4}\left(c+\frac{16}{c}\right)\ge 2.\]Adding the last three inequalities we have
\[\frac{3}{4}a+\frac{1}{2}b+\frac{1}{4}c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}\ge 8. \qquad (1)\]Using \(a+2b+3c\ge 20\) we obtain
\[\frac{1}{4}a+\frac{1}{2}b+\frac{3}{4}c\ge 5. \qquad (2)\]Finally, after adding (1) and (2) we get
\[a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}\ge 13,\]as desired.
Equality holds when \(a=2\), \(b=3\), \(c=4\), since at this point
\[a+2b+3c=2+6+12=20\]and
\[S=2+3+4+\frac{3}{2}+\frac{9}{2\cdot 3}+\frac{4}{4} =9+\frac{3}{2}+\frac{3}{2}+1 =13.\]Thus, the minimum value of \(S\) is \(13\), so \(C=13\) is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=13\).
P249. Let $a, b, c, d > 0$ be real numbers. Let $u = ab + ac + ad + bc + bd + cd$ and $v = abc + abd + acd + bcd$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c, d > 0$:
\[2u^3 \geq Cv^2\]S249. $C = 27$
We have
\[p_{2}=\frac{u}{\binom{4}{2}}=\frac{u}{6} \quad\text{and}\quad p_{3}=\frac{v}{\binom{4}{3}}=\frac{v}{4}.\]By Maclaurin’s inequality we have
\[p_{2}^{\frac{1}{2}}\ge p_{3}^{\frac{1}{3}} \ \Longleftrightarrow\ p_{2}^{3}\ge p_{3}^{2} \ \Longleftrightarrow\ \left(\frac{u}{6}\right)^{3}\ge \left(\frac{v}{4}\right)^{2} \ \Longleftrightarrow\ 2u^{3}\ge 27v^{2}.\]Equality holds when \(a=b=c=d\), that is, all variables are equal. In this case, the inequality becomes an equality, and \(C=27\) is the maximal value for which the inequality always holds.
Therefore, the answer is \(C=27\).
P250. Let $x, y, z \in \mathbb{R}^{+}$ such that $x + y + z = 1$. Find the maximal constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint:
\[\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} \geq C\]S250. $C = 1$
We have
\[\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y} =\frac{1}{2}\left(\frac{xy}{z}+\frac{yz}{x}\right) +\frac{1}{2}\left(\frac{yz}{x}+\frac{zx}{y}\right) +\frac{1}{2}\left(\frac{zx}{y}+\frac{xy}{z}\right). \qquad (1)\]Since \(AM\ge GM\) we have
\[\frac{1}{2}\left(\frac{xy}{z}+\frac{yz}{x}\right) \ge \sqrt{\frac{xy}{z}\cdot\frac{yz}{x}} =\sqrt{y^{2}} =y.\]Analogously,
\[\frac{1}{2}\left(\frac{yz}{x}+\frac{zx}{y}\right)\ge z, \qquad \frac{1}{2}\left(\frac{zx}{y}+\frac{xy}{z}\right)\ge x.\]Adding these three inequalities we obtain
\[\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\ge x+y+z=1.\]Equality holds iff \(x=y=z\), that is, when \(x=y=z=\frac{1}{3}\). In this case, the minimum value of the expression is achieved, so \(C=1\) is the maximal constant for which the inequality always holds.
Therefore, the answer is \(C=1\).
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